Question

A disc is performing pure rolling on a smooth stationary surface with constant angular velocity as shown in the figure. At any instant, for the lowermost point of the disc.

• A. Velocity is $v$ , acceleration is zero
• B. Velocity is zero, acceleration is zero
• C. Velocty is $v$ , acceleration is $\frac{v^{2}}{R}$
• D. Velocity is zero, acceleration is nonzero

Answer 1

Answer: (D)

From figure

$V_{n e t}\left(f o r \, l o w e s t \, p o i n t\right)=v-R\omega =v-v=0$
and $Acceleration=\frac{v^{2}}{R}+ \, O=\frac{v^{2}}{R}$
(Since linear speed is constant)
Hence (D)